find-minimum-value-of-f-x-4sin-2x-5sin-x-5cos-x-6-

Question Number 164716 by cortano1 last updated on 21/Jan/22

f i n d m i n i m u m v a l u e o f

f (x) = 4 \sin 2 x - 5 \sin x - 5 \cos x + 6

Answered by bobhans last updated on 21/Jan/22

f (x) = 4 \cos (\frac{π}{2} - 2 x) - 5 \sqrt{2} \cos (\frac{π}{4} - x) + 6

f (x) = 4 \cos 2 (\frac{π}{4} - x) - 5 \sqrt{2} \cos (\frac{π}{4} - x) + 6

\frac{π}{4} - x = φ \Rightarrow f (φ) = 4 \cos 2 φ - 5 \sqrt{2} \cos φ + 6

f (φ) = 8 \cos^{2} φ - 5 \sqrt{2} \cos φ + 2

f (φ) will be \min when \cos φ = \frac{5 \sqrt{2}}{16}

\min f (φ) = 8 {(\frac{5 \sqrt{2}}{16})}^{2} - 5 \sqrt{2} (\frac{5 \sqrt{2}}{16}) + 2 = \frac{7}{16}

Facebook Tweet Pin